An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.80 mm. If a 17.0 V potential difference is applied to these plates, calculate the following: ]
(a) the electric field between the plates
kV/m
(b) the capacitance
pF
(c) the charge on each plate

Explanation:

A.

V = 17.0V

d = 1.80mm

= 0.0018m

Electric field, E between the plates = V/d

= 17/0.0018

= 9444. 44 V/m.

B.

Capacitance of the capacitor, C = A ε₀/d

Where,

A = area = 7.60cm²

= 0.00076m²

d = distance between the plates = 1.80mm

= 0.0018m

ε₀ = the permittivity of free space = 8.85 x 10⁻¹²

C = 0.00076 x 8.85 x 10⁻¹² / 0.0018

= 3.74 x 10⁻¹² F.

C.

The charge on each plate, Q= C x V

Where;

V = potential difference = 17 V

C = 3.73 x 10⁻¹² F

= 3.73 x 10⁻¹² x 17

Q = 63.52 x 10⁻¹² C.

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.